Type of
Reaction
Synthesis
Indications of
Reaction
One product
Decomposition One reactant
Single
Replacement
Double
Replacement
Combustion
Reaction Formula
A + B  AB
AB  A + B
One element and A + BC  B + AC
one compound
on each side of
the arrow
Two compounds AB + BC  AD + BC
on each side of
the arrow
Oxygen is a
CxHy + O2  CO2 + H2O
reactant
Predicting
Products
Combine
elements into
compound
Break
compound into
elements
Switch metals
according to the
activity series
Switch metals
Hydrocarbons
combust to form
carbon dioxide
and water
Balancing Hints:
1. When balancing any equation start by first balancing the metals.
2. Second, balance the polyatomic ions (keeping them together helps reduce your work!)
3. Finally, balance any remaining elements.
Example:
1. Balance the following equation: Al2(SO4)3 + BaCl2  AlCl3 + BaSO4
a. Start with metals.
i. I chose to start with Aluminum. I have 2 Al on the reactant side, and 1 Al on the product side. I therefore need to multiply AlCl3 by 2 to
balance.
New Equation: Al2(SO4)3 + BaCl2  2 AlCl3 + BaSO4
ii. Next, let’s look at Barium. I have 1 Ba on the reactant side, and 1 Ba on the product side. Balanced!! 
b. Balance the polyatomic ions.
i. We have sulfate (SO42-). There are 3 sulfates on the reactant side, and 1 on the product side. This means we need to multiply the BaSO4
by 3 to balance.
New Equation: Al2(SO4)3 + BaCl2  2 AlCl3 + 3 BaSO4
ii. That now throws off our barium balance. We have 1 Ba on the reactant side and 3 Ba on the product side. So we need to multiply BaCl2
by 3 to balance.
New Equation: Al2(SO4)3 + 3 BaCl2  2 AlCl3 + 3 BaSO4
c. Balance the rest!
i. We’ve taken care of aluminum, barium, and sulfate. All that’s left is Chlorine. We have 6 Cl on the reactant side and 6 Cl on the product
side. We’re good!! (This is why we leave the individuals to the end, sometimes they work themselves out!)
Combustion Balancing Hints:
1. Start by balancing carbon.
2. Then balance hydrogen.
3. Balance oxygen last. If you get to oxygen and can’t balance it, try doubling all of your reactants (and then find the O2 coefficient).
Example:
1. Balance the following equation: C6H6 + O2  CO2 + H2O
a. Start with carbon.
i. 6 C on the reactant side, 1 on the product side. So multiply CO2 by 6 to balance.
Equation: C6H6 + O2  6 CO2 + H2O
b. Then balance hydrogen.
i. 6 H on the reactant side, 2 H on the product side. So multiply H2O by 3 to balance.
Equation: C6H6 + O2  6 CO2 + 3 H2O
c. Finish with oxygen.
i. 2 O on the reactant side, 15 on the product side.
1. This won’t work out too easily for us, so try doubling all coefficients!
Equation: 2 C6H6 + O2  12 CO2 + 6 H2O
a. Now we have still balanced C and H, and we have 2 O on the reactant side and 30 O on the product side. We can finally
balance this!!! Multiply O2 by 15.
Equation: 2 C6H6 + 15 O2  12 CO2 + 6 H2O